Question

Find the Zeroes of the given polynomials​

Answer 1

Step-by-step explanation:

We have to find the zeroes of the polynomials

$(1) {x}^{2} - 2x - 8$

Let's solve by middle term splitting method.

$= > {x }^{2} - 2x - 8 = 0 \\ \\ = > {x}^{2} + 2x - 4x - 8 = 0 \\ \\ = > x(x + 2) - 4(x + 2) = 0 \\ \\ = > (x + 2)(x - 4) = 0$

Therefore, we will get,

=> x + 2 = 0

=> x = -2

And

=> x - 4 = 0

=> x = 4

Thus, zeroes are -2 and 4.

$(2)4 {u}^{2} + 8u$

Let's solve by taking common terms

$= > 4 {u}^{2} + 8u = 0 \\ \\ = > 4u(u + 2) = 0$

Therefore, we will get,

=> 4u = 0

=> u = 0

And

=> u + 2 = 0

=> u = -2

Thus, zeroes are -2 and 0.

Answer 2

Answer:

We have to find the zeroes of the polynomials

Let's solve by middle term splitting method.

Therefore, we will get,

=> x + 2 = 0

=> x = -2

And

=> x - 4 = 0

=> x = 4

Thus, zeroes are -2 and 4.

Let's solve by taking common

terms

= > 4 {u}^{2} + 8u = 0 \\ \\ = > 4u(u + 2) = 0

Therefore, we will get,

=> 4u = 0

=> u = 0

And

=> u + 2 = 0

=> u = -2

Thus, zeroes are -2 and 0.